Question

distance of the centre of mass of a solid uniform cone from its vertex is z0 ,if the radius of its base is R and it's height h ,then Z0 is equal to​

Answer 1

We have a solid uniform right circular cone of mass M, base radius R and height H.

Right circular cone is assumed here, so the cone is symmetrical to the axis passing through its vertex and base center, and so its center of mass lies along this axis. Hence no need to find its horizontal position. We find its position along the axis, or its vertical position.

We consider an elemental cylinder of mass dM, radius r and width dh from the cone, at a height of h from the base.

In the fig., the triangles OAB and OCD are similar. Therefore,

$\sf{\longrightarrow\dfrac{H-h}{r}=\dfrac{H}{R}}$

$\sf{\longrightarrow r=\dfrac{R}{H}(H-h)\quad\quad\dots(1)}$

Volume of the cone is,

$\sf{\longrightarrow V=\dfrac{1}{3}\,\pi R^2H}$

and volume of the cylinder is,

$\sf{\longrightarrow dV=\pi r^2\ dh}$

Since the cone is of uniform, the density of both cone and cylinder are same, i.e.,

$\sf{\longrightarrow\dfrac{dM}{dV}=\dfrac{M}{V}}$

$\sf{\longrightarrow\dfrac{dM}{\pi r^2\ dh}=\dfrac{M}{\dfrac{1}{3}\,\pi R^2H}}$

$\sf{\longrightarrow\dfrac{dM}{r^2\ dh}=\dfrac{3M}{R^2H}}$

$\sf{\longrightarrow dM=\dfrac{3M}{R^2H}\ r^2\,dh}$

From (1),

$\sf{\longrightarrow dM=\dfrac{3M}{R^2H}\cdot\dfrac{R^2}{H^2}(H-h)^2\,dh}$

$\sf{\longrightarrow dM=\dfrac{3M}{H^3}(H-h)^2\,dh}$

Now the position of center of mass of the cone from the base is,

$\longrightarrow\sf{\bar h}=\dfrac{\displaystyle\sf{\int\limits_0^Mh\ dM}}{\displaystyle\sf{\int\limits_0^MdM}}$

$\longrightarrow\sf{\bar h}=\dfrac{\displaystyle\sf{\dfrac{3M}{H^3}\int\limits_0^Hh(H-h)^2\ dh}}{\displaystyle\sf{\dfrac{3M}{H^3}\int\limits_0^H(H-h)^2\ dh}}$

$\longrightarrow\sf{\bar h}=\dfrac{\displaystyle\sf{\int\limits_0^Hh(H-h)^2\ dh}}{\displaystyle\sf{\int\limits_0^H(H-h)^2\ dh}}\quad\quad\dots\sf{(2)}$

Let us evaluate each integral.

$\displaystyle\longrightarrow\sf{I_1=\int\limits_0^Hh(H-h)^2\ dh}$

$\displaystyle\longrightarrow\sf{I_1=\int\limits_0^Hh\left(H^2-2Hh+h^2\right)\ dh}$

$\displaystyle\longrightarrow\sf{I_1=\int\limits_0^H\left(H^2h-2Hh^2+h^3\right)\ dh}$

$\displaystyle\longrightarrow\sf{I_1=\dfrac{1}{2}H^2\left[h^2\right]_0^H-\dfrac{2}{3}H\left[h^3\right]_0^H+\dfrac{1}{4}\left[h^4\right]_0^H}$

$\displaystyle\longrightarrow\sf{I_1=\dfrac{1}{2}\,H^4-\dfrac{2}{3}\,H^4+\dfrac{1}{4}\,H^4}$

$\displaystyle\longrightarrow\sf{I_1=\dfrac{1}{12}\,H^4}$

And,

$\displaystyle\longrightarrow\sf{I_2=\int\limits_0^H(H-h)^2\ dh}$

$\displaystyle\longrightarrow\sf{I_2=\int\limits_0^H\left(H^2-2Hh+h^2\right)\ dh}$

$\displaystyle\longrightarrow\sf{I_2=H^2\big[h\big]_0^H-H\left[h^2\right]_0^H+\dfrac{1}{3}\left[h^3\right]_0^H}$

$\displaystyle\longrightarrow\sf{I_2=H^3-H^3+\dfrac{1}{3}\,H^3}$

$\displaystyle\longrightarrow\sf{I_2=\dfrac{1}{3}\,H^3}$

Then (2) becomes,

$\longrightarrow\sf{\bar h}=\dfrac{\sf{\left(\dfrac{1}{12}\,H^4\right)}}{\sf{\left(\dfrac{1}{3}\,H^3\right)}}$

$\sf{\longrightarrow\bar h=\dfrac{H}{4}}$

Hence the position of center of mass of the cone from its vertex will be,

$\sf{\longrightarrow z_0=H-\bar h}$

$\sf{\longrightarrow z_0=H-\dfrac{H}{4}}$

$\sf{\longrightarrow\underline{\underline{z_0=\dfrac{3H}{4}}}}$

Answer 2

$\huge{\underline{\underline{\mathrm{\red{AnswEr}}}}}$

We have a solid uniform right circular cone of mass M, base radius R and height H.

Right circular cone is assumed here, so the cone is symmetrical to the axis passing through its vertex and base center, and so its center of mass lies along this axis. Hence no need to find its horizontal position. We find its position along the axis, or its vertical position.

We consider an elemental cylinder of mass dM, radius r and width dh from the cone, at a height of h from the base.

In the fig., the triangles OAB and OCD are similar. Therefore,

$\sf{\longrightarrow\dfrac{H-h}{r}=\dfrac{H}{R}}$

$\sf{\longrightarrow r=\dfrac{R}{H}(H-h)\quad\quad\dots(1)}$

Volume of the cone is,

$\sf{\longrightarrow V=\dfrac{1}{3}\,\pi R^2H}$

and volume of the cylinder is,

$\sf{\longrightarrow dV=\pi r^2\ dh}$

Since the cone is of uniform, the density of both cone and cylinder are same, i.e.,

$\sf{\longrightarrow\dfrac{dM}{dV}=\dfrac{M}{V}}$

$\sf{\longrightarrow\dfrac{dM}{\pi r^2\ dh}=\dfrac{M}{\dfrac{1}{3}\,\pi R^2H}}$

$\sf{\longrightarrow\dfrac{dM}{r^2\ dh}=\dfrac{3M}{R^2H}}$

$\sf{\longrightarrow dM=\dfrac{3M}{R^2H}\ r^2\,dh}$

From (1),

$\sf{\longrightarrow dM=\dfrac{3M}{R^2H}\cdot\dfrac{R^2}{H^2}(H-h)^2\,dh}$

$\sf{\longrightarrow dM=\dfrac{3M}{H^3}(H-h)^2\,dh}$

Now the position of center of mass of the cone from the base is,

$\longrightarrow\sf{\bar h}=\dfrac{\displaystyle\sf{\int\limits_0^Mh\ dM}}{\displaystyle\sf{\int\limits_0^MdM}}$

$\longrightarrow\sf{\bar h}=\dfrac{\displaystyle\sf{\dfrac{3M}{H^3}\int\limits_0^Hh(H-h)^2\ dh}}{\displaystyle\sf{\dfrac{3M}{H^3}\int\limits_0^H(H-h)^2\ dh}}$

$\longrightarrow\sf{\bar h}=\dfrac{\displaystyle\sf{\int\limits_0^Hh(H-h)^2\ dh}}{\displaystyle\sf{\int\limits_0^H(H-h)^2\ dh}}\quad\quad\dots\sf{(2)}$

Let us evaluate each integral.

$\displaystyle\longrightarrow\sf{I_1=\int\limits_0^Hh(H-h)^2\ dh}$

$\displaystyle\longrightarrow\sf{I_1=\int\limits_0^Hh\left(H^2-2Hh+h^2\right)\ dh}$

$\displaystyle\longrightarrow\sf{I_1=\int\limits_0^H\left(H^2h-2Hh^2+h^3\right)\ dh}$

$\displaystyle\longrightarrow\sf{I_1=\dfrac{1}{2}H^2\left[h^2\right]_0^H-\dfrac{2}{3}H\left[h^3\right]_0^H+\dfrac{1}{4}\left[h^4\right]_0^H}$

$\displaystyle\longrightarrow\sf{I_1=\dfrac{1}{2}\,H^4-\dfrac{2}{3}\,H^4+\dfrac{1}{4}\,H^4}$

$\displaystyle\longrightarrow\sf{I_1=\dfrac{1}{12}\,H^4}$

And,

$\displaystyle\longrightarrow\sf{I_2=\int\limits_0^H(H-h)^2\ dh}$

$\displaystyle\longrightarrow\sf{I_2=\int\limits_0^H\left(H^2-2Hh+h^2\right)\ dh}$

$\displaystyle\longrightarrow\sf{I_2=H^2\big[h\big]_0^H-H\left[h^2\right]_0^H+\dfrac{1}{3}\left[h^3\right]_0^H}$

$\displaystyle\longrightarrow\sf{I_2=H^3-H^3+\dfrac{1}{3}\,H^3}$

$\displaystyle\longrightarrow\sf{I_2=\dfrac{1}{3}\,H^3}$

Then (2) becomes,

$\longrightarrow\sf{\bar h}=\dfrac{\sf{\left(\dfrac{1}{12}\,H^4\right)}}{\sf{\left(\dfrac{1}{3}\,H^3\right)}}$

$\sf{\longrightarrow\bar h=\dfrac{H}{4}}$

Hence the position of center of mass of the cone from its vertex will be,

$\sf{\longrightarrow z_0=H-\bar h}$

$\sf{\longrightarrow z_0=H-\dfrac{H}{4}}$

$\sf{\longrightarrow\underline{\underline{z_0=\dfrac{3H}{4}}}}$

${\huge{\underline{\small{\mathbb{\pink{HOPE \ THIS \ HELPED \ UH♡}}}}}}$